Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int maxn = 1e5 + 1;
const int maxj = 17;
int f[maxn][maxj], dep[maxn];
vector<int> g[maxn], a[maxn], mi[maxn][maxj];
inline void add(int x, int y) { g[x].push_back(y); }
vector<int> operator+(vector<int> a, vector<int> b) {
  for (int v : b) a.push_back(v);
  sort(a.begin(), a.end());
  while (a.size() > 10) a.pop_back();
  return a;
}
void dfs(int x, int fa) {
  f[x][0] = fa;
  dep[x] = dep[fa] + 1;
  for (int v : g[x])
    if (v != fa) {
      mi[v][0] = a[x];
      dfs(v, x);
    }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int j = maxj - 1; j >= 0; --j)
    if (dep[f[x][j]] >= dep[y]) x = f[x][j];
  if (x == y) return x;
  for (int j = maxj - 1; j >= 0; --j)
    if (f[x][j] != f[y][j]) x = f[x][j], y = f[y][j];
  return f[x][0];
}
int jump(int x, int y) {
  for (int j = maxj - 1; j >= 0; --j)
    if ((y >> j) & 1) x = f[x][j];
  return x;
}
vector<int> get(int x, int p) {
  vector<int> ret = a[x];
  for (int j = maxj - 1; j >= 0; --j)
    if (dep[f[x][j]] >= dep[p]) ret = ret + mi[x][j], x = f[x][j];
  return ret;
}
int main() {
  int n = read(), m = read(), q = read();
  for (int i = 1; i < n; ++i) {
    int x = read(), y = read();
    add(x, y), add(y, x);
  }
  for (int i = 1; i <= m; ++i) a[read()].push_back(i);
  for (int i = 1; i <= n; ++i) {
    sort(a[i].begin(), a[i].end());
    while (a[i].size() > 10) a[i].pop_back();
  }
  dfs(1, 1);
  for (int j = 1; j < maxj; ++j)
    for (int i = 1; i <= n; ++i)
      f[i][j] = f[f[i][j - 1]][j - 1],
      mi[i][j] = mi[i][j - 1] + mi[f[i][j - 1]][j - 1];
  while (q--) {
    int x = read(), y = read(), s = read();
    if (dep[x] > dep[y]) swap(x, y);
    int l = lca(x, y);
    int p = jump(y, dep[y] - dep[l] - 1);
    vector<int> ans = (x == l ? get(y, l) : get(x, l) + get(y, p));
    while (ans.size() > s) ans.pop_back();
    printf("%d ", ans.size());
    for (int v : ans) printf("%d ", v);
    puts("");
  }
  return 0;
}